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zavuch27 [327]
3 years ago
6

What are the mean, median, and mode of the data in the following sample?

Mathematics
1 answer:
Leto [7]3 years ago
6 0

Answer:

A.

Step-by-step explanation: You have to add all the numbers together for the mean then for the median you put them in order from least to greatest. Then put your finger on the first and last number and start counting down until you have the middle number.  

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The environment club has saved 200 pounds of cans and promises to collect 20 pounds of cans each week and will use the profits t
anzhelika [568]

Answer:

20

Step-by-step explanation:

Since they are collecting 20 pounds of cans each week, the slope is 20/1, or 20.

Hope this helps!

5 0
3 years ago
Read 2 more answers
Figure EFGHK is to be transformed to Figure E'F'G'H'K' using the rule (x, y)→(x + 8, y + 5):
IceJOKER [234]
The rule that has been given means that x coordinate of some point we increase by 8 and y coordinate of some point we increase by 5.

Coordinates of K point on original Figure are:
(-2,-3)

once we implement rule on this we get K':

K' ( -2+8,-3+5) 
or 
K' ( 6,2)

Answer is third option.

8 0
3 years ago
Read 2 more answers
Please don’t skip!!! I need help!!
horsena [70]
P = [60] cm

Explanation:

6.5 cm + 6.5 cm + 7.5 cm + 7.5 cm + 7.5 cm + 7.5 cm + 8.5 cm + 8.5 cm
7 0
3 years ago
Please Help !!Mr. Mudd gives each of his children $2000 to invest as part of a friendly family competition. The competition will
VikaD [51]

Answer:

Albert = $2159.07; Marie = $2244.99; Hans = $2188.35; Max = $2147.40

Marie is $10 000 richer

Step-by-step explanation:

Albert

(a) $1000 at 1.2 % compounded monthly

A = P\left(1 + \dfrac{r}{n}\right)^{nt}

A = 1000(1 + 0.001)¹²⁰ = $1127.43

(b) $500 losing 2%

0.98 × 500 = $490

(c) $500 compounded continuously at 0.8%

\begin{array}{rcl}A & = & Pe^{rt}\\& = & 500e^{0.008 \times 10}\\& = &\mathbf{\$541.64}\\\end{array}\\

(d) Balance

Total = 1127.43 + 490.00+ 541.64 = $2159.07

Marie

(a) 1500 at 1.4 % compounded quarterly

A = 1500(1 + 0.0035)⁴⁰ = $1724.99

(b) $500 gaining 4 %

1.04 × 500 = $520.00

(c) Balance

Total = 1724.99 + 520.00 = $2244.99

Hans

$2000 compounded continuously at 0.9 %

\begin{array}{rcl}A& = &2000e^{0.009 \times 10}\\& = &\mathbf{\$2188.35}\\\end{array}\\

Max

(a) $1000 decreasing exponentially at 0.5 % annually

A = 1000(1 - 0.005)¹⁰= $951.11

(b) $1000 at 1.8 % compounded biannually

A = 1000(1 + 0.009)²⁰ = $1196.29

(c) Balance

Total = 951.11 + 1196.29 = $2147.40

Marie is $ 10 000 richer at the end of the competition.

7 0
4 years ago
A highway department executive claims that the number of fatal accidents which occur in her state does not vary from month to mo
den301095 [7]

This question is incomplete, the complete question;

A highway department executive claims that the number of fatal accidents which occur in her state does not vary from month to month. The results of a study of 193 fatal accidents were recorded. Is there enough evidence to reject the highway department executive's claim about the distribution of fatal accidents between each month? Month Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec Fatal Accidents 11 13 22 11 15 29 15 9 13 15 22 18

Step 10 of 10 : State the conclusion of the hypothesis test at the 0.1 level of significance.

Answer: H₀ : The distribution of fatal accidents between each month is same

Step-by-step explanation:

H₀ : The distribution of fatal accidents between each month is same

H₁ : The distribution of fatal accidents between each month is different

Let the los be alpha 0.10

given data ;

       Observed    Expected  

Mon   Freq (Oi)   Freq Ei       (Oi-Ei)^2 /Ei

Jan      11    16.0833        1.606649

Feb      13    16.0833        0.591105

Mar      22    16.0833        2.176598

Apr       11    16.0833        1.606649

May     15    16.0833        0.072971

Jun     29    16.0833        10.373489

Jul     15    16.0833        0.072971

Aug       9    16.0833        3.119603

Sep      13    16.0833        0.591105

Oct      15    16.0833        0.072971

Nov      22    16.0833        2.176598

Dec      18    16.0833       0.228411

Total:   193     93                22.689119

Expected Freq ⇒ 193 / 12 = 16.08333

Test Statistic, x² : 22.6891

Num Categories: 12

Degrees of freedom: 12 - 1 = 11

Critical value X² : 24.725

P-Value: 0.0195

since Chi-square value < Chi-square critical value

and P-value > alpha 0.01 so we accept H₀

Therefore we conclude that the distribution of fatal accidents between each month is same

7 0
3 years ago
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